Problem: Find the sum of the first $50$ terms in this geometric series: $1 + \dfrac{10}{11} + \dfrac{100}{121}...$ Choose 1 answer: Choose 1 answer: (Choice A) A $0.01$ (Choice B) B $0.99$ (Choice C) C $10.91$ (Choice D) D $11$
Getting started We're dealing with a geometric series because each term is multiplied by $\dfrac{10}{11}$ to get the next term. We need a formula to compute the sum of the terms. Formula for geometric series The sum $S_n$ of a finite geometric series is $S_n = \dfrac{a_1(1-r^n)}{1-r}$ where $a_1$ is the first term, $r$ is the common ratio, and $n$ is the number of terms. What do we need to use the formula? The first term $(a_1 = {1})$ and the number of terms $(n = {50})$ are given in the question. The common ratio $r$ is ${\dfrac{10}{11}}$ because each term is multiplied by ${\dfrac{10}{11}}$ to get the next term. [How did we find the common ratio r?] Find the sum $(S_n)$ of the series $\begin{aligned} S_n &= \dfrac{a_1(1-r^n)}{1-r} \\\\ S_{{50}}&=\dfrac{{1}\left(1-\left({\dfrac{10}{11}}\right)^{{50}}\right)}{1-{\dfrac{10}{11}}} \\\\ S_{{50}}&=11\left(1-\left({\dfrac{10}{11}}\right)^{{50}}\right)\\\\ S_{{{50}}} &\approx 10.91\end{aligned}$ The answer $10.91$